B.2.2.16 IVP of \dfrac{x(x^2+1)}{4y^3}, y(0)=-\dfrac{1}{\sqrt{2}}

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In summary: Mahalo for your help.In summary, the solution to the initial value problem $y^{\prime} = \dfrac{x(x^2+1)}{4y^3}, \quad y(0)=-\dfrac{1}{\sqrt{2}}$ is $y = -\sqrt{\dfrac{x^2+1}{2}}$ on the interval $(-\infty, \infty)$.
  • #1
karush
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Find the solution of $y^{\prime} = \dfrac{x(x^2+1)}{4y^3}, \quad y(0)=-\dfrac{1}{\sqrt{2}}$
in explicit form.

rewrite $\dfrac{y}{x}=\dfrac{x(x^2+1)}{4y^3}= y^3\, dy=\dfrac{x(x^2+1)}{4}\, dx$
integrate $\dfrac{y^4}{4}= \dfrac{1}{4}\left(\dfrac{x^4}{4} +\dfrac{ x^2}{2}\right)$
$y^4=\dfrac{x^4}{4} +\dfrac{ x^2}{2}+C$
$y=\left[\dfrac{x^4}{4} +\dfrac{ x^2}{2}+ C\right]^{1/4}$
obtain C $y(0)=\left[\dfrac{0}{4} +\dfrac{ 0}{2}+ C\right]^{1/4}=-\dfrac{1}{\sqrt{2}}
\implies =C^{1/4}=-\dfrac{1}{\sqrt{2}}
=C=-\left(\dfrac{1}{\sqrt{2}}\right)^4=-\dfrac{1}{4}$
finally $y=-\left[\dfrac{x^4}{4} +\dfrac{ x^2}{2}+\dfrac{1}{4}\right]^{1/4}
=-\left[ \dfrac{(x^2+ 1)^2}{4}\right]^{1/4}$
(b) Plot the graph of the solution.
(c)Determine the interval of the solution
book answer $y=−\sqrt{\dfrac{(x^2+1}{2}} \quad [−\infty<x<\infty]$
ok i think there are some bugs in this and probably some suggestions
I wanted to plot this in tikz but did know how to deal with the final eq
mahalo
 
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  • #2
Are you sure you've copied the problem correctly?

Mathematica is showing no solution. The problem is likely with the initial condition.
 
  • #3
karush said:
Find the solution of $y^{\prime} = \dfrac{x(x^2+1)}{4y^3}, \quad y(0)=-\dfrac{1}{\sqrt{2}}$
in explicit form.

rewrite $\dfrac{y}{x}=\dfrac{x(x^2+1)}{4y^3}= y^3\, dy=\dfrac{x(x^2+1)}{4}\, dx$
What is this line? You can't possibly think that this is correct? I used to take points off for lines like this.

Write it out!
\(\displaystyle \dfrac{dy}{dx} = \dfrac{x(x^2 + 1)}{4y^3}\)

\(\displaystyle y^3 ~ dy = \dfrac{x(x^2 + 1)}{4} ~ dx\)

-Dan
 
  • #4
romsek said:
Are you sure you've copied the problem correctly?

Mathematica is showing no solution. The problem is likely with the initial condition.

Screenshot 2021-09-09 10.56.38 AM.png

here is the book answer
Screenshot 2021-09-09 11.08.27 AM.png
 
Last edited:
  • #5
topsquark said:
What is this line? You can't possibly think that this is correct? I used to take points off for lines like this.

Write it out!
\(\displaystyle \dfrac{dy}{dx} = \dfrac{x(x^2 + 1)}{4y^3}\)

\(\displaystyle y^3 ~ dy = \dfrac{x(x^2 + 1)}{4} ~ dx\)

-Dan
ok it should of been and $\implies$ or a new line
 
  • #6
We easily obtain that

\(\displaystyle \dfrac{dy}{dx} = \dfrac{x(x^2+1)}{4y^3}\\

4 y^3 dy = (x^3 + x) dx\\

y^4 = \dfrac 1 4 \left(x^4 + 2 x^2 + C\right)\\~\\

\text{It's easiest to evaluate the constant here}\\

(y(0))^4 = \dfrac 1 4 C = \left(-\dfrac{1}{\sqrt{2}}\right)^4 = \dfrac 1 4 \Rightarrow C = 1\\

y =\pm \left(\dfrac 1 4 \left(x^4 + 2x^2 + 1\right)\right)^{1/4} = \\

\pm \dfrac{1}{\sqrt{2}}\left(\left(x^2+1\right)^2\right)^{1/4} = \\

\pm \dfrac{1}{\sqrt{2}}\sqrt{(x^2+1)}\\

\text{and by the initial condition we see that we only keep the negative solution, i.e.}\\

y(x) = -\dfrac{1}{\sqrt{2}}\sqrt{x^2+1} = -\sqrt{\dfrac{x^2+1}{2}}
\)
 
  • #7
Mahalo
That helped a lot
 

FAQ: B.2.2.16 IVP of \dfrac{x(x^2+1)}{4y^3}, y(0)=-\dfrac{1}{\sqrt{2}}

What does "IVP" stand for in this equation?

"IVP" stands for "initial value problem". It is a type of mathematical problem that involves finding a solution to a differential equation that satisfies given initial conditions.

What is the significance of the value of y(0)=-1/√2 in this equation?

The value of y(0)=-1/√2 is the initial condition given in the problem. It represents the value of y at the initial time or starting point of the problem. In this case, it is used to help find the particular solution to the differential equation.

How do you solve for the solution to this initial value problem?

To solve this initial value problem, you would use the method of separation of variables. This involves isolating the variables on opposite sides of the equation and then integrating both sides to find the solution.

What is the domain and range of the solution to this initial value problem?

The domain of the solution is all real numbers except for 0, since the denominator cannot be equal to 0. The range of the solution is also all real numbers, since the function is continuous and has no restrictions on its output.

Can you provide a real-life example of how this initial value problem may be used?

This initial value problem can be used to model the growth or decay of a population over time, where x represents time and y represents the size of the population. The initial condition y(0)=-1/√2 could represent the initial population size at time 0, and the solution to the problem would give the population size at any given time.

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